Physics MCQs

Question #601

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The root mean square speed of gas molecules is proportional to:

A. Temperature

B. Square root of temperature

C. Square of temperature

D. Inverse of temperature

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Correct Answer: Option B

Explanation:
From kinetic theory, rms speed v_rms = √(3RT/M), where R is gas constant, T absolute temperature, M molar mass. Thus v_rms ∝ √T. This shows molecular speed increases with temperature but not linearly. Memory aid: 'Kinetic energy ∝ T; KE = ½mv² ⇒ v ∝ √T'. This conceptual question tests kinetic theory understanding, frequently examined in thermodynamics sections of competitive exams. Always use absolute temperature (Kelvin) in gas law calculations; Celsius would give incorrect proportionality.

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Question #602

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In a potentiometer experiment, the balancing length for a cell is 60 cm. When the cell is shunted by a 2Ω resistor, the balancing length becomes 40 cm. The internal resistance of the cell is:

A. 2Ω

B. 3Ω

C. 1Ω

D. 4Ω

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Correct Answer: Option C

Explanation:
Potentiometer measures EMF and internal resistance. Formula: r = R(L₁ - L₂)/L₂, where R is shunt resistance, L₁ initial balancing length, L₂ after shunting. Here R=2Ω, L₁=60 cm, L₂=40 cm. Thus r = 2×(60-40)/40 = 2×20/40 = 1Ω. This derives from terminal voltage V = E - Ir and potentiometer principle V ∝ balancing length. Memory tip: 'r = R(L₁/L₂ - 1)'. This advanced numerical tests potentiometer applications, common in competitive exams for experimental physics. Always verify the formula derivation to avoid misapplication.

This question belongs to: Science Physics

Question #603

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A body is projected with velocity u at angle θ to horizontal. The horizontal range is maximum when θ is:

A. 45°

B. 60°

C. 90°

D. 30°

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Correct Answer: Option A

Explanation:
Horizontal range R = u²sin(2θ)/g. sin(2θ) is maximum (value 1) when 2θ = 90° ⇒ θ = 45°. Thus maximum range at 45° projection angle, assuming same initial speed and level ground. Memory tip: 'Max range at 45°; max height at 90°'. This standard projectile motion result is frequently tested in competitive exams. Always verify assumptions: no air resistance, uniform gravity, launch and landing at same height. For different heights, optimal angle differs.

This question belongs to: Science Physics

Question #604

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The critical angle for light passing from glass (n=1.5) to air is approximately:

A. 90°

B. 42°

C. 30°

D. 60°

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Correct Answer: Option B

Explanation:
Critical angle θ_c for total internal reflection: sinθ_c = n₂/n₁, where n₁ > n₂. Here n₁=1.5 (glass), n₂=1 (air), so sinθ_c = 1/1.5 = 2/3 ≈ 0.6667 ⇒ θ_c = sin⁻¹(0.6667) ≈ 41.8° ≈ 42°. Memory aid: 'sinθ_c = 1/n for glass to air'. This calculation tests understanding of total internal reflection conditions, crucial for optics in competitive exams. Always ensure light travels from denser to rarer medium for TIR to be possible; otherwise, no critical angle exists.

This question belongs to: Science Physics

Question #605

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In Young's double slit experiment, if the slit separation is doubled and screen distance halved, the fringe width becomes:

A. Double

B. Half

C. Same

D. One-fourth

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Correct Answer: Option D

Explanation:
Fringe width β = λD/d, where λ wavelength, D screen distance, d slit separation. New d' = 2d, new D' = D/2. Thus β' = λ(D/2)/(2d) = λD/(4d) = β/4. Fringe width reduces to one-fourth. Memory tip: 'β ∝ D/d; changes multiply'. This proportional reasoning problem tests wave optics understanding, frequently appearing in competitive exams. Always track how each parameter change affects the result; combine multiplicative factors for net effect. Verify with dimensional analysis: β has length dimension, consistent with λD/d.

This question belongs to: Science Physics

Question #606

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The work function of a metal is 2.3 eV. The threshold frequency for photoelectric emission is approximately: (h = 4.14×10⁻¹⁵ eV·s)

A. 8.0×10¹⁴ Hz

B. 5.6×10¹⁴ Hz

C. 3.5×10¹⁴ Hz

D. 1.2×10¹⁵ Hz

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Correct Answer: Option B

Explanation:
Photoelectric equation: work function φ = hν₀, where ν₀ is threshold frequency. Thus ν₀ = φ/h = 2.3 eV / (4.14×10⁻¹⁵ eV·s) ≈ 5.555×10¹⁴ Hz ≈ 5.6×10¹⁴ Hz. This direct application tests photoelectric effect fundamentals. Memory aid: 'ν₀ = φ/h; higher work function ⇒ higher threshold frequency'. Competitive exams frequently provide h in eV·s for such calculations. Always ensure units match: eV for φ and h to avoid conversion errors. This problem assesses numerical proficiency in modern physics applications.

This question belongs to: Science Physics

Question #607

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A Carnot engine operates between 400 K and 300 K. Its efficiency is:

A. 75%

B. 25%

C. 20%

D. 33.3%

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Correct Answer: Option B

Explanation:
Carnot efficiency η = 1 - T₂/T₁, where T₁ is source temperature, T₂ sink temperature (in Kelvin). Here T₁=400 K, T₂=300 K, so η = 1 - 300/400 = 1 - 0.75 = 0.25 = 25%. This maximum possible efficiency for given temperatures is a fundamental thermodynamics result. Memory tip: 'η_Carnot = 1 - T_cold/T_hot'. Competitive exams frequently test this formula with varying temperatures. Always use absolute temperatures (Kelvin); Celsius would yield incorrect efficiency. This problem assesses understanding of heat engine limitations.

This question belongs to: Science Physics

Question #608

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The electric field due to an infinite plane sheet of charge with surface charge density σ is:

A. 2σ/ε₀

B. σ/(2ε₀)

C. Zero

D. σ/ε₀

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Correct Answer: Option B

Explanation:
Using Gauss's law, for an infinite sheet, electric field E = σ/(2ε₀), directed perpendicular to the sheet. This is uniform and independent of distance from the sheet. Option A is for field near conductor surface; C and D are incorrect. Memory aid: 'Infinite sheet: E = σ/(2ε₀); conductor surface: E = σ/ε₀'. This conceptual question tests electrostatics fundamentals, crucial for competitive exams. Always distinguish between conducting and non-conacting sheets, as field expressions differ due to charge distribution.

This question belongs to: Science Physics

Question #609

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A capacitor of capacitance C is charged to voltage V. The energy stored is:

A. CV²

B. ½CV²

C. ½C/V

D. CV

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Correct Answer: Option B

Explanation:
Energy stored in capacitor U = ½CV² = ½QV = Q²/(2C). This derives from work done to charge the capacitor against increasing voltage. Option A has wrong dimensions (energy should be joules, CV is coulomb-volt = joule, but missing factor ½); C and D are dimensionally incorrect. Memory tip: 'Capacitor energy = ½CV², analogous to spring energy ½kx²'. This formula-based question tests electrostatics knowledge, frequently appearing in competitive exams. Always recall the three equivalent forms and choose based on given quantities.

This question belongs to: Science Physics

Question #610

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The magnetic susceptibility of a diamagnetic material is:

A. Large and positive

B. Small and positive

C. Large and negative

D. Small and negative

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Correct Answer: Option D

Explanation:
Diamagnetic materials have small negative susceptibility (χ ≈ -10⁻⁵), meaning they weakly repel magnetic fields. Paramagnetic: small positive χ; ferromagnetic: large positive χ. This classification is based on response to external magnetic fields. Memory aid: 'Diamagnetic: repelled by magnets; χ < 0'. This conceptual question tests magnetism fundamentals, crucial for competitive exams. Always link susceptibility sign to material behavior: negative χ ⇒ diamagnetic (weak repulsion), positive ⇒ paramagnetic/ferromagnetic (attraction).

This question belongs to: Science Physics

Question #611

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In a transistor, the current gain β is defined as:

A. I_C / I_B

B. I_C / I_E

C. I_B / I_C

D. I_E / I_C

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Correct Answer: Option A

Explanation:
For a bipolar junction transistor (BJT) in common-emitter configuration, DC current gain β = I_C / I_B, where I_C is collector current, I_B base current. Typical β values range 20-200. Option D is α (common-base gain). Memory tip: 'β = collector current / base current; large β means small base current controls large collector current'. This definition-based question tests electronics fundamentals, frequently examined in competitive exams. Always distinguish β (CE configuration) from α (CB configuration) and their relationship: β = α/(1-α).

This question belongs to: Science Physics

Question #612

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The binding energy per nucleon is maximum for:

A. Helium

B. Hydrogen

C. Iron

D. Uranium

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Correct Answer: Option C

Explanation:
Binding energy per nucleon peaks around iron-56 (≈8.8 MeV/nucleon), making it the most stable nucleus. Lighter nuclei release energy via fusion; heavier via fission. Hydrogen has near-zero binding energy per nucleon; uranium ≈7.6 MeV/nucleon. Memory aid: 'Iron peak: most stable nucleus; fusion before, fission after'. This nuclear physics concept is frequently tested in competitive exams. Always link binding energy curve to energy release in stars (fusion) and reactors (fission).

This question belongs to: Science Physics

Question #613

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The time period of oscillation of a mass M attached to a spring of force constant k is:

A. π√(M/k)

B. 2π√(k/M)

C. 2π√(M/k)

D. 2π√(g/l)

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Correct Answer: Option C

Explanation:
For spring-mass system undergoing SHM, time period T = 2π√(M/k), derived from F = -kx = Ma ⇒ a = -(k/M)x, so ω² = k/M, T=2π/ω=2π√(M/k). Option B has inverse ratio; C misses factor 2; D is for pendulum. Memory tip: 'Spring: T ∝ √(M/k); Pendulum: T ∝ √(l/g)'. This standard formula is frequently tested in oscillations sections of competitive exams. Always verify dimensions: √(kg / (N/m)) = √(kg·m/N) = √(s²) = s, correct for time period.

This question belongs to: Science Physics

Question #614

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A ray of light is incident on a glass slab at angle i. The emergent ray is:

A. Parallel to incident ray but laterally displaced

B. Bent towards the normal

C. Bent away from the normal

D. Perpendicular to incident ray

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Correct Answer: Option A

Explanation:
In a parallel-sided glass slab, refraction at first surface bends ray towards normal; at second surface (glass to air), it bends away by equal angle, making emergent ray parallel to incident ray but shifted laterally. No net deviation, only displacement. Memory aid: 'Slab: emergent ray parallel to incident; prism: emergent ray deviated'. This ray optics concept is frequently tested in competitive exams. Always distinguish slab (parallel faces) from prism (non-parallel faces) behavior: slab causes lateral shift, prism causes angular deviation.

This question belongs to: Science Physics

Question #615

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The resolving power of a microscope is increased by:

A. Reducing the magnification

B. Using oil immersion

C. Decreasing the numerical aperture

D. Using light of longer wavelength

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Correct Answer: Option B

Explanation:
Resolving power (minimum resolvable distance) d = λ/(2NA), where NA is numerical aperture. Oil immersion increases NA by matching refractive index between slide and objective, reducing light refraction and increasing light collection. Longer wavelength (A) decreases resolving power; decreasing NA (C) worsens resolution; magnification (D) doesn't affect fundamental resolution limit. Memory aid: 'Higher NA or shorter λ ⇒ better resolution'. This application question tests optical instrument knowledge, common in competitive exams. Always link resolution to wave nature of light (diffraction limit) and instrument design factors.

This question belongs to: Science Physics

Question #616

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The phenomenon of beats in sound waves is due to:

A. Diffraction

B. Interference

C. Reflection

D. Refraction

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Correct Answer: Option B

Explanation:
Beats occur when two sound waves of slightly different frequencies interfere, causing periodic variation in amplitude (loudness). Beat frequency = |f₁ - f₂|. This is a superposition/interference phenomenon. Reflection, refraction, diffraction don't produce beats. Memory tip: 'Beats = interference of close frequencies; used in tuning instruments'. This conceptual question tests wave superposition understanding, frequently examined in competitive exams. Always distinguish beats (amplitude modulation) from resonance (amplitude increase at natural frequency).

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Question #617

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The dimensional formula of angular momentum is:

A. [ML²T⁻¹]

B. [M⁰L⁰T⁰]

C. [MLT⁻¹]

D. [ML²T⁻²]

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Correct Answer: Option A

Explanation:
Angular momentum L = r × p. Position r has [L], linear momentum p = mv has [MLT⁻¹], so L has [L][MLT⁻¹] = [ML²T⁻¹]. This matches Planck's constant dimensions. Option A is linear momentum; C is energy/torque; D is dimensionless. Memory aid: 'Angular momentum: [ML²T⁻¹]; same as action quantities'. This dimensional analysis question tests ability to derive formulas, crucial for competitive exams. Always break down compound quantities into fundamental dimensions (M, L, T) for verification and problem-solving.

This question belongs to: Science Physics

Question #618

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A body is thrown vertically upwards. At the highest point of its trajectory:

A. Velocity is zero but acceleration is g downward

B. Velocity and acceleration are zero

C. Velocity is maximum and acceleration is zero

D. Both velocity and acceleration are maximum

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Correct Answer: Option A

Explanation:
At maximum height, vertical velocity becomes zero momentarily. However, acceleration due to gravity (g) acts downward throughout the motion, including at the peak. Thus acceleration is g ≈ 9.8 m/s² downward, not zero. Memory tip: 'Peak of projectile: v=0, a=g downward'. This conceptual question tests understanding of motion under gravity, frequently appearing in competitive exams. Always distinguish velocity (which changes) from acceleration (constant in free fall near Earth). Misconception that acceleration is zero at peak is a common trap.

This question belongs to: Science Physics

Question #619

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The coefficient of performance of a refrigerator is defined as:

A. Work input / Heat extracted

B. Heat extracted / Work input

C. Heat rejected / Work input

D. Work input / Heat rejected

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Correct Answer: Option B

Explanation:
Coefficient of performance (COP) for refrigerator: COP = Q₂ / W, where Q₂ is heat extracted from cold reservoir, W is work input. By energy conservation, Q₁ = Q₂ + W (heat rejected to hot reservoir). COP > 1 typically. Option A is inverse; C and D relate to heat engine efficiency. Memory aid: 'Refrigerator COP = desired output (cooling) / work input'. This thermodynamics definition is frequently tested in competitive exams. Always distinguish refrigerator COP (cooling effect per work) from heat engine efficiency (work output per heat input).

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Question #620

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In electromagnetic waves, the electric and magnetic fields are:

A. Parallel to each other but perpendicular to propagation

B. Perpendicular to each other and to direction of propagation

C. Randomly oriented

D. Parallel to each other and to direction of propagation

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Correct Answer: Option B

Explanation:
Electromagnetic waves are transverse: electric field (E) and magnetic field (B) are perpendicular to each other and both perpendicular to the direction of wave propagation. This is a fundamental property derived from Maxwell's equations. Memory tip: 'EM waves: E ⊥ B ⊥ propagation direction; right-hand rule for orientation'. This conceptual question tests electromagnetic wave fundamentals, crucial for modern physics in competitive exams. Always recall that EM waves require no medium and travel at speed of light in vacuum.

This question belongs to: Science Physics

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